Chapter 13.2, Problem 8E

Weekly Earnings of Women According to the Women’s Bureau of the U.S. Department of Labor, the occupation with the highest median weekly earnings among women is pharmacist with median weekly earnings of $1603. Based on the weekly earnings listed from a random sample of female pharmacists, can it be concluded that the median is less than $1603? Use α = 0.05.

1550	1355	1777
1430	1570	1701
2465	1655	1484
1429	1829	1812
1217	1501	1449

To determine

To identify: The claim and state $H_0$ and $H_1$.

Answer to Problem 8E

The claim is “the median weekly earnings for U.S. women pharmacist is 1603 dollars”.

Null hypothesis: $H_0:\text{Median}=1,603$

Alternative hypothesis: $H_1:\text{Median}<1,603$

Explanation of Solution

Given info:

The data shows the number of female pharmacist is 15.

Justification:

Here, the claim is that median weekly earnings for U.S. women pharmacist is lesser than 1603 dollars. This can be written as $\text{Median}<1,603$. The complement of the claim is $\text{Median}=1,603$. In the experiment, the alternative hypothesis indicates the claim.

To determine

To find: The critical value.

Answer to Problem 8E

The critical value is 3.

Explanation of Solution

Determine sample size n:

Subtract the hypothesized median of 1,603 from the data.

Data	$\text{Difference}=\text{Data}-1,603$	Sign	Data	Diff	Sign
1550	$1,550–1,603=–53$	–	1484	$1,484–1,603=–119$	–
1355	$1,355–1,603=–248$	–	1429	$1,429–1,603=–174$	–
1777	$1,777–1,603=174$	+	1829	$1,829–1,603=226$	+
1430	$1,430–1,603=–173$	–	1812	$1,812–1,603=209$	+
1570	$1,570–1,603=–33$	–	1217	$1,217–1,603=–386$	–
1701	$1,701–1,603=98$	+	1501	$1,501–1,603=–102$	–
2465	$2,465–1,603=862$	+	1449	$1,449–1,603=–154$	–
1655	$1,655–1,603=52$	+

The sample size is,

$\begin{array}{c}n=\text{sum of }(+)\text{ and }(-)\text{ signs}\\ =6+9\\ =15\end{array}$

Critical value:

From table” critical values for sign test” for $n=15$ at $\alpha =0.05$.

The critical value is 3.

To determine

To find: the test statistic $x$

Answer to Problem 8E

The test statistic is 6.

Explanation of Solution

Condition:

If $n\le 25,$ the test statistic for sign test is $x$.

If $n>25$, the test statistic for sign test is $z=\frac{\left(x+0.5\right)-0.5n}{\frac{\sqrt{n}}{2}}$

Where $x$ is the smaller number of $(+)$& $(-)$ and $n$ is the sample size.

Here, the number of $(+)$ sign in the table is 6 and number of $(-)$ sign in the table is 9.

Thus the test statistic is 6.

To determine

To decide: Whether to reject or fail to reject the null hypothesis at level of significance $\alpha =0.05$.

Answer to Problem 8E

The decision is “fail to reject the null hypothesis”.

Explanation of Solution

Decision rule:

If the test statistic is less than or equal to the critical value, then reject the null hypothesis $H_0$.

If the test statistic is greater than the critical value, then fail to reject the null hypothesis $H_0$.

Conclusion:

The level of significance is, $\alpha =0.05$ and the critical value is 3.

Here, the test statistic value is greater than the critical value.

That is, $\text{Test statistic}\left(=6\right)>\text{criticalvalue}\left(=3\right)$

Thus, the decision is “fail to reject the null hypotheses”.

To determine

To interpret: The decision in the context of the original claim.

Answer to Problem 8E

The conclusion is that there is enough evidence to reject the claim that the median weekly earnings for U.S. women pharmacist are less than 1603 dollars.

Explanation of Solution

From the part (d), the null hypothesis is not rejected. Thus, there is enough evidence to reject the claim that the median weekly earnings for U.S. women pharmacist are less than $1,603 at $5\%$ level of significance.